# nLab partial trace

Suppose $V$, $W$ are finite-dimensional vector spaces over a field, with dimensions $m$ and $n$, respectively. For any space $A$ let $L\left(A\right)$ denote the space of linear operators on $A$. The partial trace over $W$, Tr${}_{W}$, is a mapping

$T\in L\left(V\otimes W\right)↦{\mathrm{Tr}}_{W}\left(T\right)\in L\left(V\right).$T \in L(V \otimes W) \mapsto Tr_{W}(T) \in L(V).
###### Definition

Let ${e}_{1},\dots ,{e}_{m}$ and ${f}_{1},\dots ,{f}_{n}$ be bases for $V$ and $W$ respectively. Then $T$ has a matrix representation $\left\{{a}_{\mathrm{kl},\mathrm{ij}}\right\}$ where $1\le k,i\le m$ and $1\le l,j\le n$ relative to the basis of the space $V\otimes W$ given by ${e}_{k}\otimes {f}_{l}$. Consider the sum

${b}_{k,i}=\sum _{j=1}^{n}{a}_{\mathrm{kj},\mathrm{ij}}$b_{k,i} = \sum_{j=1}^{n}a_{kj,ij}

for $k,i$ over $1,\dots ,m$. This gives the matrix ${b}_{k,i}$. The associated linear operator on $V$ is independent of the choice of bases and is defined as the partial trace.

# Example

Consider a quantum system, $\rho$, in the presence of an environment, ${\rho }_{\mathrm{env}}$. Consider what is known in quantum information theory as the CNOT gate:

$U=\mid 00⟩⟨00\mid +\mid 01⟩⟨01\mid +\mid 11⟩⟨10\mid +\mid 10⟩⟨11\mid .$U={|00\rangle}{\langle 00|} + {|01\rangle}{\langle 01|} + {|11\rangle}{\langle 10|} + {|10\rangle}{\langle 11|}.

Suppose our system has the simple state $\mid 1⟩⟨1\mid$ and the environment has the simple state $\mid 0⟩⟨0\mid$. Then $\rho \otimes {\rho }_{\mathrm{env}}=\mid 10⟩⟨10\mid$. In the quantum operation formalism we have

$T\left(\rho \right)=\frac{1}{2}{\mathrm{Tr}}_{\mathrm{env}}U\left(\rho \otimes {\rho }_{\mathrm{env}}\right){U}^{†}=\frac{1}{2}{\mathrm{Tr}}_{\mathrm{env}}\left(\mid 10⟩⟨10\mid +\mid 11⟩⟨11\mid \right)=\frac{\mid 1⟩⟨1\mid ⟨0\mid 0⟩+\mid 1⟩⟨1\mid ⟨1\mid 1⟩}{2}=\mid 1⟩⟨1\mid$T(\rho) = \frac{1}{2}Tr_{env}U(\rho \otimes \rho_{env})U^{\dagger} = \frac{1}{2}Tr_{env}({|10\rangle}{\langle 10|} + {|11\rangle}{\langle 11|}) = \frac{{|1\rangle}{\langle 1|}{\langle 0|0\rangle} + {|1\rangle}{\langle 1|}{\langle 1|1\rangle}}{2} = {|1\rangle}{\langle 1|}

where we inserted the normalization factor $\frac{1}{2}$.

Revised on March 6, 2010 23:40:54 by Toby Bartels (75.117.106.207)